Lesson video

Hello, I'm Mr. Langton, and today we're going to be looking at using tree diagrams for combined events.

All you're going to need is something to write with or something to write on.

It might be useful to have a calculator for some of the trickier calculations.

Try to make sure you're in a quiet space with no distractions.

And when you're ready, we'll begin.

We'll start with the, "try this" activity.

One cube is drawn from each of the three bags shown and then replaced.

Xavier and Yasmin are trying to work out the probability of drawing the same colour from all three bags.

Xavier says, I'm going to draw a sample space for the three combined events." Yasmin thinks, it would be easier, would be better to draw a tree diagram.

Who do you agree with, and why? Pause the video, have a think can make some notes.

When you're ready, unpause it, and we'll go through it together.

You can pause in three, two, one.

So how did you get on? I'm going to start by looking at Xavier's idea.

And I'll start it by drawing a sample space for the first 2 bags.

Bag A, has got one green and white cubes.

Bag B, has got three greens and two whites.

So, draw on a sample space to show that? And the problem comes when I introduce this third bag, because I don't know now where I'm going to draw my next part of my sample space.

The moment we've got a third event happening, it's very, very difficult to draw.

Technically, you could try and expand it to make a 3D model, instead.

That's going to get very confusing, and it's going to be impossible to try to draw on in the fourth dimension if we have a fourth event happening.

This is where tree diagrams come in come in, really handy.

So, let's have a look at the tree diagram.

We'll start with bag A.

It's got one green cube and three white cubes.

So, when we draw a tree diagram, it's going to look like that.

The probability of getting green, is going to be 1/4, probability of getting white, will be 3/4.

Now, let's look at bag B.

There are three green cubes, the probability of get in green is 3/5.

And there are two white cubes, the probability of getting white is 2/5.

Now, presume we drew a white one first, we then got to fill this in, again.

The probability of getting green, is 3/5, and the probability of getting white is 2/5.

Okay.

So far, we can now introduce that third bag.

So, we get to bag C, and there are now three greens out of six which I'm going to write as 1/2.

And then also three whites out of six, so it's a half.

And I've got to do that every time, every possible event.

So, let's look at all the possible outcomes that could happen.

If we follow the lines all the way along to the top, here is the probability, that get three greens in a row.

Here is the probability, that I get two greens followed by a white.

Then we have a probability that I get green, and then a white, and then a green.

We've got a green followed by white, followed by a white.

We've got the probability of choosing a white followed by a green and then a green.

White, green, white white, white, green, and white, white, white.

So, representing the tree diagram, we managed to show all the possible options that we can get.

Now that we're trying to find out the probability of choosing the same colour each time, so that's either going to be three greens, which we got at the top, or, three whites, which is at the bottom.

We can work out those probabilities now.

The probability.

I can do that in green line.

The probability of choosing three greens, will be 1/4 the first bag, multiplied by 3/5 of the second bag, multiplied by 1/2 for the third bag, which is going to be three over.

4 x 5 = 20, 3/40.

And I don't know how to write with white on white background, So I'm going to use red for this one.

The probability of getting three whites is going to be 3/4 in the first bag, 2/5 in the second bag, and 1/2 in the third bag.

We're multiplying those together 3 x 2 x 1 = 6, 4 x 5 = 20 x 2 = 40.

I've got the same denominator in each case, the probability that I get to have the same colour, can be 3/40 + 6/40 = 9/40.

Okay.

Now it's your turn.

This question involves the decimal instead of a fraction, and because we're multiplying it three times, it's going to give me quite a tricky answer.

So this is where you might want to use a calculator to work out your answer.

Pause the video, and have a go, when you're ready, unpause it, and we'll go through it together.

Good luck.

How did it get on? I've completed the tree diagram for you.

On the right hand side, I put all the possible outcomes as a list.

Let's have the first one done.

Part B, what is the probability that it lands on heads every time? There's only one way that can happen, that's that top one.

We need to do 0.

7 x 0.

7 x 0.

7, And that's where you're going to want your calculator because it's 0.

343.

I wouldn't expect you to do that in your head.

I'm sure you could work it out with pen and paper, But it'd be easy with a calculator.

Part C.

What is the probability that it lands on tails at least once? So let's go for green that's got at least one tail.

That one does, that one does, that one does, that one does, that one does, and that one does.

Now, it can be a lot of work to work on all of those calculations and add them all together.

But did you notice that it's every single outcome apart from the first one? We've already worked out first one, since we know that all of the probabilities in order to make a whole one, if we subtract the one that we don't need, then, we're going to get our answer.

So that's the probability that lands on tails at least once.

Finally, what is the probability that it lands on heads at least twice? So it's that's that one, that one, that one, and that one.

The probability of getting three heads, we've already worked out, is 0.

343.

Now the probability of getting two heads and a tail, that's going to be 0.

7 x 0.

7 x 0.

3, which your calculator should tell you is 0.

147.

So, I've done that one, and I've done that one.

Now heads, tails, heads.

I'm doing the same calculation because multiplication is commutative into 0.

7 x 0.

3 x 0.

7.

So once again, 0.

147.

And we've done it once more there and again it's two heads and a tail, so it's going to be the same answer, 0.

147.

So all we need to do, is add these three together.

I'm going to do that on a calculator, now, giving me 0.

784.

And that's the final answer.

We'll finish with the Explorer activity.

There are 10 white and 10 green cubes and they're being shared into three bags.

One cube is then drawn from each bag and replaced.

Xavier and Yasmin are discussing the probability of drawing all three cubes the same colour.

Xavier says, It doesn't matter how we arrange the cubes in the bag, the probability will always be the same.

Yasmin says, well, it does matter because the probability will change depending on how they're share it out.

Who do you agree with? And can you explain why? Pause the video, make some notes, have a go When you're ready, unpause it and we'll discuss it together.

You can pause in three, two, one.

So, how did you get on? This one stopped me for a little bit.

I had to think about it for quite long.

Then, I thought, well, it does matter how they're arranged.

Just for example, with all the green cubes, ending up in the first bag, no white cube is going in the first bag.

That would make it absolutely impossible to draw three cubes of the same Colour, But I could have a different outcome where I've got at least two greens in each bag, at least two whites in each bag.

So, there, it's definitely possible that it can happen.

So, I might come with an example where the two probabilities are definitely different.

So I think Xavier is wrong.

You asked me is correct.

The probability has changed depending on how the cubes are shared out.

We'd love to see how you'd be getting on with the work.

If you'd like to share it with us, please ask your parents or curator to share your work on Twitter, tagging "@OakNational" and "#LearnwithOak" Thank you very much.

Goodbye.